Question

A silvery white metal on treatment with NaOH and HCI liberates hydrogen gas to form B and respectively. The metal A will not react with acid D due to the formation passive film on the surface. Hence it is used for transporting acid D. Identify A, B, C, D and support the answer with balanced equations.​

Answer 1

Answer:

A=Al

Al+NaOH-->NaAlO2+H2

Al+HCl-->AlCl3+H2

B=AlCl3

C=NaAlO2

D=HNO3

Answer 2

Answer:

The silvery-white metal that reacts with both acids and bases is Al.

2 Al + 6 HCl ⟶ 2 AlCl₃ + 3 H₂

2 Al + 2 NaOH + 6 H₂O ⟶ 2Na⁺ + [Al(OH)₄]⁻ + 3H₂

However Al does not react with concentrated HNO₃ due to formation of passive oxide layer on the surface.

Hence, A = Al, B = Na[Al(OH)₄] , C = AlCl₃ , D = HNO₃ (concentrated)  

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